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Thread: finding length

  1. #1
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    Question finding length

    A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?
    Last edited by zak_killcity14@hotmail.co; Dec 1st 2006 at 06:50 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by zak_killcity14@hotmail.co View Post
    A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?
    the figure is like this:
    make a right angled triangle, right angled at D
    label one vertex as A, other as B
    On DB, choose a point C
    Now, angleABD(or ABC)=45degrees
    and angleACD=67degrees
    BC=32 m
    you have to find AD
    from triangle ACD
    $\displaystyle \frac{AD}{CD}=tan67$
    or$\displaystyle CD=\frac{AD}{tan67}......(1)$
    from triangle ABD
    $\displaystyle \frac{AD}{BD}=tan45$
    or$\displaystyle BD=\frac{AD}{tan45}......(2)$
    Subtracting (1) from (2)
    $\displaystyle BD-CD=\frac{AD}{tan45} - \frac{AD}{tan67}$
    but BD-CD=BC=32
    hence
    $\displaystyle 32=\frac{AD}{tan45} - \frac{AD}{tan67}$
    $\displaystyle AD=$$\displaystyle \frac{32}{\frac{1}{tan45} - \frac{1}{tan67}}$

    Keep Smiling
    Malay
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  3. #3
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    Hello, Zak!

    I assume you have a sketch . . .


    A steel chimney is held in position by several guy wires.
    Two such wires $\displaystyle AB$ and $\displaystyle AC$ where $\displaystyle B,\,C$ and the base of the chimney, $\displaystyle D$,
    are on the same level in a straight line with $\displaystyle B$ and $\displaystyle C$ on the left of the chimney.
    If $\displaystyle \text{angle }ABC = 45^o,\text{ angle }ACD = 67^o,\;BC = 32\text{ m,}$
    find the height of the chimney, $\displaystyle AD.$
    Code:
                                  * A
                               * *|
                            *   * |
                         *     *  |
                      *       *   | y
                   *         *    |
                *           *     |
             * 45         * 67  |
          *---------------*-------*
          B      32       C   x   D

    Let $\displaystyle x = CD,\:y = AD.$

    In right triangle $\displaystyle ADB:\;\tan45^o = \frac{y}{x+32}\quad\Rightarrow\quad1 = \frac{y}{x+32}\quad\Rightarrow\quad x \:= \:y - 32$ [1]

    In right triangle $\displaystyle ADC:\;\tan67^o = \frac{y}{x}\quad\Rightarrow\quad x \,=\,\frac{y}{\tan67^o}$ [2]


    Equate [1] and [2]: .$\displaystyle y - 32 \:=\:\frac{y}{\tan67^o}$


    Solve for $\displaystyle y:\;\;y\tan67^o - 32\tan67^o \:=\:y\quad\Rightarrow\quad y\tan67^o - 32\tan67^o \:=\:y$

    . . $\displaystyle y\tan67^o - y \:=\:32\tan67^o\quad \Rightarrow\quad y(\tan67^o - 1) \:=\:32\tan67^o$


    Therefore: .$\displaystyle y \;=\;\frac{32\tan67^o}{\tan67^o - 1} \;=\;55.60138965 \;\approx\;\boxed{55.6\text{ m}}$

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